Q. 8.18 - A First Course in Probability

Question

Repeat part $\left(a\right)$ of Problem$8.2$when it is known that the variance of a student’s test score is equal

to$25$.

Step-by-Step Solution

Answer

Therefore,

$P\{X\ge 85\}\le \frac{1}{5}=.2$.

Step 1 Given Information.

the variance of a student’s test score is equal to$25$.

Step 2 Explanation.

Let $X$represents the test score of a student taking her final examination. Assume that $X$is a random variable with mean $\mu =75$and variance${\sigma }^2$$=25$.

Then, using Corollary$5.1$. (textbook)$\underset{¯}{a=10}$, we get:

$P\{X\ge 85\}=P\{X\ge 75+10\}\le \frac{{\sigma }^2}{{\sigma }^2+{10}^2}=\frac{25}{25+100}=\frac{1}{5}=0.2$